PAT A1099 Build A Binary Search Tree （30 分）——二叉搜索树，中序遍历，层序遍历

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

Given the structure of a binary tree and a sequence of distinct integer keys, there is only one way to fill these keys into the tree so that the resulting tree satisfies the definition of a BST. You are supposed to output the level order traversal sequence of that tree. The sample is illustrated by Figure 1 and 2.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤100) which is the total number of nodes in the tree. The next N lines each contains the left and the right children of a node in the format left_index right_index, provided that the nodes are numbered from 0 to N−1, and 0 is always the root. If one child is missing, then −1 will represent the NULL child pointer. Finally N distinct integer keys are given in the last line.

Output Specification:

For each test case, print in one line the level order traversal sequence of that tree. All the numbers must be separated by a space, with no extra space at the end of the line.

Sample Input:

9
1 6
2 3
-1 -1
-1 4
5 -1
-1 -1
7 -1
-1 8
-1 -1
73 45 11 58 82 25 67 38 42

Sample Output:

58 25 82 11 38 67 45 73 42

 

 1 #include 
 2 #include 
 3 #include <set>
 4 #include 
 5 #include 
 6 using namespace std;
 7 const int maxn = 110;
 8 int n;
 9 struct node{
10     int data;
11     int l=-1,r=-1;
12 }bst[maxn];
13 int index=0;
14 int q[maxn];
15 void inorder(int root){
16     if(bst[root].l!=-1) inorder(bst[root].l);
17     bst[root].data = q[index++];
18     if(bst[root].r!=-1) inorder(bst[root].r);
19 }
20 void level(int root){
21     queue<int> que;
22     int cnt=0;
23     que.push(root);
24     while(!que.empty()){
25         int now=que.front();
26         que.pop();
27         printf("%d",bst[now].data);
28         cnt++;
29         if(cnt!=n) printf(" ");
30         if(bst[now].l!=-1) que.push(bst[now].l);
31         if(bst[now].r!=-1) que.push(bst[now].r);
32     }
33 }
34 int main(){
35     scanf("%d",&n);
36     for(int i=0;i){
37         int l,r;
38         scanf("%d %d",&l,&r);
39         bst[i].l=l;
40         bst[i].r=r;
41     }
42     for(int i=0;i){
43         scanf("%d",&q[i]);
44     }
45     sort(q,q+n);
46     inorder(0);
47     level(0);
48 }

View Code

注意点：其实很简单的题目，又一次倒在了递归上。知道要把给定数据sort，但没想sort完后就是这棵树的中序遍历结果，只要把正常中序遍历时的打印改成赋值就能得到那棵树了。没想到中序遍历，所以自己一直在想怎么把这个有序序列一个个填到树里去，一直也没搞明白，看这题通过率0.5多，我还不会做，凉了。

看到树的题目，一般都会要建树，遍历，而对bst，一般都是要对给定序列排序的，这样能得到他的中序遍历结果，有助于建树

转载于:https://www.cnblogs.com/tccbj/p/10447363.html